 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-3/15d/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Parametrize the loop by

r^{'} = R \overset{e}{^}_{r} (ϕ)

and

d ℓ^{'} = R \overset{e}{^}_{ϕ} (ϕ) d ϕ

. At the observation point

r = z \overset{e}{^}_{z}

d B = \frac{μ _{0} I}{4 π} \frac{d ℓ ^{'} \times ( r - r ^{'} )}{( R ^{2} + z ^{2} ) ^{3/2}} .

The components perpendicular to the axis cancel by rotational symmetry, while the axial component of the cross product is

R^{2} d ϕ

. Therefore

B (z) = \frac{μ _{0} I R ^{2}}{2 ( R ^{2} + z ^{2} ) ^{3/2}} \overset{e}{^}_{z} .

The direction is fixed by the right-hand rule and reverses when the electric current reverses. At the centre,

B (0) = \frac{μ _{0} I}{2 R} \overset{e}{^}_{z} .

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